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800 ⏐⏐⏐ NETWORK THEOREMS (ac)
+
E
–
Z 1
Z 2 ETh
FIG. 18.26
Determining the open-circuit Thévenin
voltage for the network of Fig. 18.23.
+
ETh = 3.33 V ∠ – 180°
–
+
–
Z Th = 2.67 � ∠ –90°
Z Th
�j 16 �
ZTh � � � ��
6 �90°
� 2.67 � ��90°
Step 4 (Fig. 18.26):
Z2E ETh � � (voltage divider rule)
Z1 � Z2 2 Z1Z2 ( j 8 �)(�j 2 �) 16 �
� �� �
Z1 � Z2 j 8 ��j 2 � j 6
(�j 2 �)(10 V) �j 20 V
��� ���3.33 V ��180°
j 8 ��j 2 � j 6
Step 5: The Thévenin equivalent circuit is shown in Fig. 18.27.
R
FIG. 18.27
The Thévenin equivalent circuit for the network of Fig. 18.23.
+
X C = 2.67 �
Th
ETh = 3.33 V ∠ – 180° R
–
EXAMPLE 18.8 Find the Thévenin equivalent circuit for the network
external to branch a-a′ in Fig. 18.28.
E 1
+
–
R 1
6 �
10 V ∠ 0°
X L1
8 �
X C
R 2
3 �
X L2 = 5 �
4 �
FIG. 18.28
Example 18.8.
a
R 3
7 �
+
E 2
–
a� Thévenin
30 V ∠ 15°
Solution:
Steps 1 and 2 (Fig. 18.29): Note the reduced complexity with subscripted
impedances:
+
E 1
–
10 V ∠ 0°
Z 1
Z 2
Z 3
a
a� Thévenin
FIG. 18.29
Assigning the subscripted impedances to the network of Fig. 18.28.